Question

In proof testing of circuit boards, the probability that anyparticular diode will fail is 0.01. Suppose...

In proof testing of circuit boards, the probability that anyparticular diode will fail is 0.01.
Suppose a circuit board contains 200 diodes. How manydiodes would you expect to fail, and what is the standarddeviation of the number that are expected to fail? What is the (approximate) probability that atleast four diodes will fail on a randomly selected board?If five boards are shipped to a particularcustomer, how likely is it that at least four of them willwork properly? (A board works properly only if all its diodes work)
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Concepts and reason

Get the binomial distribution under the following experimental conditions:

(i)Each trial results in two exhaustive and mutually disjoint outcomes, termed as success and failure.

(ii)The number of trials n is finite.

(iii)The trials are independent of each other.

(iv)The probability of success p is constant for each trial.

Binomial distribution is important not only because of its wide applicability, but because it gives rise to many other probability distributions.

Fundamentals

Let the random variable X follows a binomial distribution with parameters n and p.

The probability mass function (PMF) of binomial distribution can be defined as,

--ar
Р(х -х)-
х%3D0,1, 2, ..., п
(1-р);
х

The formula for the mean of the binomial distribution is,

и3 Е (x)
%3 пр

The formula for the standard deviation of the binomial distribution is,

РX)
- пр (1- р)
(х)
о3

From the given information we have n 200
and p 0.01

Let X denotes number of diode fail based on a circuit board contains 200 diodes

Here, X Binom(n 200, p = 0.01)

The mean and standard deviation of the probability distribution of number of diode fail based on a circuit board contains 200 diodes are,

(200)(0.01)
=2

Vnp(1-p)
= 200(0.01)(1-0.01)
=1.407125
1.407

The objective is to find that the probability that at least four diodes will fail on a randomly selected board. Suppose that a circuit board contains 200 diodes.

From the given information,n=200, p 0.01

Let X denotes number of diode fail based on a circuit board contains 200 diodes

Here,

Then, the probability function for the given binomial random variable is,

200
(0.01) (0,99)
200
P(X x)
x 0,1,2,200

Compute the probability that at least four diodes will fail on a randomly selected board.

P(X24) 1-P(X<4)
SP(X 0)+P(X )
1+P(x=2)+ P(X =3)
200
0.01 (09)
1-
200
200-0
200-
J(001(0.99)
0
=1-
200
(0.01 (0.99)*
200
(0.

Compute the probability that randomly selected board have 0 diodes fail.

200
(0.01)(0.99)*
200-0
P(X=0)=
200
=1)((0,99)0
0.1340

That is, probability that a randomly selected circuit board work properly is 0.1340.

Suppose five boards are shipped to a particular customer.

Let Y denotes number circuit boards work properly.

Here, Y Binom(m 5, p 0.1340)

The probability function for the binomial random variable Y with parameters 5 and 0.1340 is,

P(Y y)
(0.134) (1-0.134)y0,1,2,3,4,5

Compute the probability that at least four of the boards will work properly.

P(Y2 4) P(Y 4)+P(Y =5)
|(0.134) (1-0.134)
((0.134) (1-0.134)
0.0014+0.00004
=0.00144

Ans:

On average we expect 2 diodes expect to fail and the standard deviation of the number that are expected to fail is 1.407.

The probability that at least four diodes will fail on a randomly selected board is 0.142.

Probability that at least four of the boards will work properly is 0.00144.

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